Question

A random sample of 22 students’ weights is drawn from student population. Investigate whether the average weight of student population is different from 140 lb. 135 119 106 135 180 108 128 160 143 175 170 205 195 185 182 150 175 190 180 195 220 235 State the null and alternative hypothesis (Ho and Ha). What are the n,X ̅, s? Compute the t-statistic. What is the degree of freedom (df)? Find P-value from the table-D. Test the hypothesis at the significance level α=0.05. Reject Ho or Ha? Why? What conclusion can you make about the mean weight of students? Construct 95% Confidence Interval for the students’ mean weight. What value for t* should you use? Find the t* value from table-D. Do the calculation for the 95% Confidence Interval. Based on the 95% Confidence Interval from i), what conclusion can you make about the hypothesis in a)? Why?

(Answer all the questions and in a word document plz)

Answer 1

n = 22

= (135 + 119 + 106 + 135 + 180 + 128 + 160 + 143 + 175 + 170 + 205 + 195 + 185 + 182 + 150 + 175 + 190 + 180 + 195 + 220 + 235)/22 = 161.955

s = sqrt(((135 - 161.955)^2 + (119 - 161.955)^2 + (106 - 161.955)^2 + (135 - 161.955)^2 + (180 - 161.955)^2 + (128 - 161.955)^2 + (160 - 161.955)^2 + (143 - 161.955)^2 + (175 - 161.955)^2 + (170 - 161.955)^2 + (205 - 161.955)^2 + (195 - 161.955)^2 + (185 - 161.955)^2 + (182 - 161.955)^2 + (150 - 161.955)^2 + (175 - 161.955)^2 + (190 - 161.955)^2 + (180 - 161.955)^2 + (195 - 161.955)^2 + (220 - 161.955)^2 + (235 - 161.955)^2)/21) = 33.527

  

df = 22 - 1 = 21

P-value = 2 * P(T > 3.071)

= 2 * (1 - P(T < 3.071))

= 2 * (1 - 0.9971)

= 2 * 0.0029

= 0.0058

The P-value < alpha = 0.05. So reject H0.

At alpha = 0.05, we can can conclde that the average weight of student population is different from 140 lb.

The critical value t* =

The 95% confidence interval for population mean is

Since the population mean 140 does not lie in the confidence interval, so we should reject H0.